# homology of $\mathbb{RP}^{3}$.

We need for this problem knowledge of the homology groups of $S^{2}$ and $\mathbb{RP}^{2}$. We will simply assume the former:

 $\displaystyle H_{k}(S^{2};\mathbb{Z})$ $\displaystyle=\begin{cases}\mathbb{Z}&k=0,2\\ 0&else\end{cases}$

Now, for $\mathbb{RP}^{2}$, we can argue without Mayer-Vietoris. $X=\mathbb{RP}^{2}$ is connected, so $H_{0}(X;\mathbb{Z})=\mathbb{Z}$. $X$ is non-orientable, so $H_{2}(X;\mathbb{Z})$ is 0. Last, $H_{1}(X;\mathbb{Z})$ is the abelianization of the already abelian fundamental group $\pi_{1}(X)=\mathbb{Z}/2\mathbb{Z}$, so we have:

 $\displaystyle H_{k}(\mathbb{RP}^{2};\mathbb{Z})$ $\displaystyle=\begin{cases}\mathbb{Z}&k=0\\ \mathbb{Z}/2\mathbb{Z}&k=1\\ 0&k\geq 2\end{cases}$

Now that we have the homology of $\mathbb{RP}^{2}$, we can compute the homology of $\mathbb{RP}^{3}$ from Mayer-Vietoris. Let $X=\mathbb{RP}^{3}$, $V=\mathbb{RP}^{3}\backslash\{pt\}\sim\mathbb{RP}^{2}$ (by vieweing $\mathbb{RP}^{3}$ as a CW-complex), $U\sim D^{3}\sim\{pt\}$, and $U\cap V\sim S^{2}$, where $\sim$ denotes equivalence through a deformation retract. Then the Mayer-Vietoris sequence gives